If $ x+4 y=14 $ is a normal to the curve $ y^{2}=\alpha x^{3}-\beta $ at $ (2,3) $, then value of $ \alpha+\beta $ is :

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

x+4 y=14

is a normal to the curve

y^{2}=\alpha x^{3}-\beta

(2,3)

, then value of

\alpha+\beta

is :

View answer and explanation

Correct option: A

Option A (9) is correct because substituting the point (2,3) into the curve equation and using the slope of the normal line allows us to solve for alpha and beta, resulting in alpha + beta = 9.

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